Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to solve this system of equations? $$\begin{cases} 1+\sqrt{2 x+y+1}=4 (2 x+y)^2+\sqrt{6 x+3 y},\\ (x+1) \sqrt{2 x^2-x+4}+8 x^2+4 x y=4. \end{cases}$$

share|cite|improve this question
up vote 2 down vote accepted

The trick here is to not use both equations at the same time.

The first equation screams to use the substitution $k=2x+y$. Substituting that, we get $1+\sqrt{k+1}=4k^2+\sqrt{3k}$. So we transform it to $(4k^2-1)+(\sqrt{3k}-\sqrt{k+1})=0$. So $(4k^2-1)(\sqrt{3k}+\sqrt{k+1})+(2k-1)=0$.

(this is because we multiplied both sides by $\sqrt{3k}+\sqrt{k+1}$.

Thus we have either $2k-1=0$ or $(2k+1)(\sqrt{3k}+\sqrt{k+1})+1=0$. For the quantity $(2k+1)(\sqrt{3k}+\sqrt{k+1})$ to be real, we need $k$ nonnegative. However, if $k$ is nonnegative, then the quantity is clearly positive, so no roots.

Thus we need $k=\frac{1}{2}$. $8x^2+4xy=4x(2x+y)=2x$. So $(x+1)\sqrt{2x^2-x+4}+2x=4$. $(x+1)\sqrt{2x^2-x+4}=4-2x$, and squaring we get $(x+1)^2(2x^2-x+4)=(4-2x)^2$. After expanding, we get $2x^4+3x^3+23x-12=0$, so it factors as $(2x-1)(x+3)(x^2-x+4)=0$. Since the discriminant of $x^2-x+4$, which is $1-4(4)$, is negative, it doesn't have a real root.

Hence the possible values of $x$ are $\frac{1}{2}$ and $-3$, giving values of $y$ as $\frac{-1}{2}$ and $\frac{-11}{2}$. We could just "check" them to see that both work. (just plug back in)

share|cite|improve this answer

You have two equations with two variables. $$1+\sqrt{2x+y+1}-4(2x+y)^2-\sqrt{6x+3y}=0$$ $$(x+1)\sqrt{2x^2-x+4}+8x^2+4xy-4=0$$ Solve it with any root finding algortihm.

If you are looking for real solutions $$x=0.5\qquad y=-0.5$$

share|cite|improve this answer

Hint Define $U=2x+y$ in first equation

You will get $1+\sqrt{U+1}=4 U^2+\sqrt{3U}$ solve $U$

here you need to solve $U$

$$(x+1) \sqrt{2 x^2-x+4}+8 x^2+4xy=4$$

$$(x+1) \sqrt{2 x^2-x+4}+4x(2x+y)=4$$

Then put U in second equation and find $x$

$$(x+1) \sqrt{2 x^2-x+4}+4xU=4$$

after finding $x$ , you can get $y$ from $U=2x+y$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.